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Question
Form the pair of linear equation in the following problem, and find its solution (if they exist) by the elimination method:
If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes `1/2` if we only add 1 to the denominator. What is the fraction?
Solution
Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is `x/y`
If 1 is added to the numerator and 1 is subtracted from the denominator, the fraction becomes 1. Thus, we have
`(x + 1)/(x - y)=1`
⇒ x + 1 = y - 1
⇒ x - y = -2 ...(i)
If 1 is added to the denominator, the fraction becomes `1/2`. Thus, we have
`x/(y + 1)=1/2`
`x = 1/2 (y + 1)`
`x - y/2 = 1/2` ...(ii)
By subtracting (2) from (1) we have
`x - y - x + y/2=-2 - 1/2`
⇒ `-y + y/2=-2 - 1/2`
⇒ `-1/2y = -5/2`
⇒ y = 5
Now, putting y = 5 in (ii), we have
`x - 5/2 = 1/2`
`x = 1/2 + 5/2`
`x=6/2`
x = 3
Thus, x = 3 and y = 5
Hence, the required fraction = `3/5`
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